Capacitors with unequally charged plates?

I'm currently on the chapter on capacitors in my physics book. I'm starting to get the concept, but everything still isn't crystal clear. I get that you can store charges in two plates, one containing a charge of Q, and the other a charge of -Q. But, would it not also work if the charges where unequal to each other in magnitude? Say, if you have one plate with +Q charge and another with -2Q ? Sorry if this is a really obvious question, haha.

Staff: Mentor

Sure, you can store different amounts of charges on each plate, but you probably won't find any real world circuits that do that. Capacitors are placed into circuits uncharged and neutral, and the circuits aren't usually capable of charging each plate separately.

If you connect a capacitor to a battery and then put it all on top of a Van der Graaf ball (say, positively charged), there will be an exess of positive charge on both plates of your capacitor. Hence the two pates will have different net charges on them, the positive plate having more and the negative plate having fewer.
Mostly, because the capacity between capacitor plates is much higher than the capacity of the individual plates to Earth and the voltages are all 'low', there will be a very small value of unbalanced charge. But you could say that your suggestion nearly always applies - just not to any significant degree.
In RF circuitry, which could involve C values of only a few pF, the imbalance could be much more significant (although the values of actual charge would be tiny).

one thing to get clear in your thinking is that if you connect a capacitor to a battery the charge on each plate MUST be equal. The charge removed from one plate is placed on the other plate... so equal.
If you have 2 plate with different charges it cannot be a capacitor connected to a battery.

The capacitor in the OP was not connected to a battery, though. But I think I am still right under those circumstances. Charge the capacitor with +1C on one terminal and -1C on the other, then charge each plate (independently) with another +0.001C (involving a lot of volts, of course). The net result will be +1.001C on one and -0.999C on the other.
I don't think the addition of a battery should alter that.